题目内容
已知数列{an}的前n项和Sn满足
=
(a>0),数列{bn}满足bn=an•logaan
(1)求数列{an}的通项;
(2)求数列{bn}的前n项和Tn.
| an-1 |
| Sn |
| a-1 |
| a |
(1)求数列{an}的通项;
(2)求数列{bn}的前n项和Tn.
(1)当n=1时,a1=a>0且a≠1,
当n≥2时,Sn=
(an-1),Sn-1=
(an-1-1),
两式相减得an=
(an-an-1),化为
=a,
∴数列{an}是等比数列,an=an;
(2)bn=an•lo
=nan.
当a=1时,Tn=1+2+…+n=
.
当a≠1时,Tn=a+2a2+3a3+…+nan,
aTn=a2+2a3+…++(n-1)an+nan+1,
∴(1-a)Tn=a+a2+…+an-nan+1=
-nan+1,
∴Tn=
.
当n≥2时,Sn=
| a |
| a-1 |
| a |
| a-1 |
两式相减得an=
| a |
| a-1 |
| an |
| an-1 |
∴数列{an}是等比数列,an=an;
(2)bn=an•lo
| g | ana |
当a=1时,Tn=1+2+…+n=
| n(1+n) |
| 2 |
当a≠1时,Tn=a+2a2+3a3+…+nan,
aTn=a2+2a3+…++(n-1)an+nan+1,
∴(1-a)Tn=a+a2+…+an-nan+1=
| a(an-1) |
| a-1 |
∴Tn=
| a+(na-n-1)an+1 |
| (a-1)2 |
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